Chemistry, 21.04.2020 15:27 suyi14

How many milliliters of 2.0M Mg(OH)2 are required to neutralize 250.0 ml of 0.80M hydrochloric acid (HCl)?

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Answer from: doggosbepis

OK, This is my thinking, I hope this helps you out.

Let's do problem "a"

Step 1. Write a balanced equation.

2 HCl + Ba(OH)2 > BaCl2 + 2 HOH

Step 2. Underneath the HCl and Ba(OH)2 in the equation write what you are given and what you need to find out.

2 HCl + Ba(OH)2 > BaCl2 + 2 HOH

C = 0.130 mol/L v = 56 mL

V = ? C = 0.109 mol/L

Step 3. They ALWAYS provide a way to calculate the number of moles of one of the substances. In this case it is Ba(OH)2 because you have C and V

n = C X V n = 0.109 x 56 = 6.104 millimoles

Step 4. Use the equation to figure out how many moles of the unknown (HCl) you will need.

the equation says you need TWICE as many moles as you have of Ba(OH)2. So you will need 6.104 millimoles x 2 or 12.208 milimoles.

Step 5. Now that yo have moles and concentration of HCl you can now calculate volume

C = n / V, so V = n / C

12.208 millimoles / 0.130 mol/L = 93.9 mL

Use this method for all these kind of problems.

In B) you will have to convert he g of NaOH into moles by 0.240 / 40.00 g/mol and then following trhe rest of the procedure.

C) is done the same way. convert the g of Na2SO4 into moles first.

d) is almost the same as a) except this time when you find the moles of Ca(OH)2 you just have to convert the moles into mass by multiplying moles x Molar mass of Ca(OH)2.

Best of luck.

Explanation:

Answer from: Quest

a is the answer to this question

explanation:

Answer from: Quest

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1

0.44 / 0.09 = 4.8

mgco3-5h2o"

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How many milliliters of 2.0M Mg(OH)2 are required to neutralize 250.0 ml of 0.80M hydrochloric acid...